1. **State the problem:** We have an equilateral triangle $\triangle ABC$ with side length 5 cm. An arc $BDC$ is drawn with center at $A$ and radius equal to $AB=5$ cm. We need to find the area of the segment $BCD$ formed by the arc and chord $BC$. 2. **Understand the geometry:** Since $\triangle ABC$ is equilateral, all sides are 5 cm and all angles are $60^\circ$. The arc $BDC$ is part of a circle centered at $A$ with radius 5 cm. The chord $BC$ subtends an angle of $60^\circ$ at $A$. 3. **Formula for the area of a segment:** The area of a segment formed by a chord and an arc is: $$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$ 4. **Calculate the area of sector $BAC$:** The sector angle is $60^\circ$ or $\frac{\pi}{3}$ radians. Radius $r = 5$ cm. $$\text{Area of sector} = \frac{\theta}{2\pi} \times \pi r^2 = \frac{\theta}{2} r^2$$ Substitute $\theta = \frac{\pi}{3}$: $$\text{Area of sector} = \frac{\pi/3}{2} \times 5^2 = \frac{\pi}{6} \times 25 = \frac{25\pi}{6}$$ 5. **Calculate the area of triangle $ABC$:** Equilateral triangle area formula: $$\text{Area} = \frac{\sqrt{3}}{4} s^2$$ Substitute $s=5$: $$\text{Area} = \frac{\sqrt{3}}{4} \times 25 = \frac{25\sqrt{3}}{4}$$ 6. **Calculate the area of segment $BCD$:** $$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle} = \frac{25\pi}{6} - \frac{25\sqrt{3}}{4}$$ 7. **Final answer:** $$\boxed{\text{Area of segment } BCD = \frac{25\pi}{6} - \frac{25\sqrt{3}}{4} \text{ cm}^2}$$ This is the exact area of the segment formed by the arc $BDC$ and chord $BC$ in the equilateral triangle.